hdoj--1789--Doing Homework again(贪心模拟)

最新推荐文章于 2020-04-02 15:59:51 发布

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最新推荐文章于 2020-04-02 15:59:51 发布

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本文探讨了学生如何在给定的截止日期下完成作业，以最小化最终成绩的减分。通过贪心算法，优先完成分数较高的作业，并确保在截止日期之前完成，以实现整体减分最小化。

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Doing Homework again

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10441 Accepted Submission(s): 6113

Problem Description

Ignatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every teacher gives him a deadline of handing in the homework. If Ignatius hands in the homework after the deadline, the teacher will reduce his score of the final test. And now we assume that doing everyone homework always takes one day. So Ignatius wants you to help him to arrange the order of doing homework to minimize the reduced score.

Input

The input contains several test cases. The first line of the input is a single integer T that is the number of test cases. T test cases follow.
Each test case start with a positive integer N(1<=N<=1000) which indicate the number of homework.. Then 2 lines follow. The first line contains N integers that indicate the deadlines of the subjects, and the next line contains N integers that indicate the reduced scores.

Output

For each test case, you should output the smallest total reduced score, one line per test case.

Sample Input

  
   3
3
3 3 3
10 5 1
3
1 3 1
6 2 3
7
1 4 6 4 2 4 3
3 2 1 7 6 5 4

Sample Output

Author

lcy

Source

2007省赛集训队练习赛（10）_以此感谢DOOMIII

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lcy | We have carefully selected several similar problems for you: 1978 2159 1159 1069 2571

还是完成作业那一类的题目，每一个作业对应一定的学分，还有一个最后上交时间，求最少的损失

就是一个贪心，最大的分值肯定不能丢，但是太早完成又会耽搁其他的作业，所以就拖到最后完成，然后就是第二大分值的，一个一个贪心，无法完成是因为他没有前边的分值大，并且略微靠前，我们前边的都已经是最后的交稿时间，所以不能牺牲前边的完成后边小的

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm> 
using namespace std;
struct node
{
	int d,val;
}p[10100];
int n,vis[10010];
bool cmp(node s1,node s2)
{
	return s1.val>s2.val;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(int i=0;i<n;i++)
		scanf("%d",&p[i].d);
		for(int i=0;i<n;i++)
		scanf("%d",&p[i].val);
		memset(vis,0,sizeof(vis));
		sort(p,p+n,cmp);
		__int64 sum=0;
		for(int i=0;i<n;i++)
		{
			int k=0;
			for(int j=p[i].d;j>0;j--)
			{
				if(!vis[j])
				{
					vis[j]=1;
					k=1;
					break;
				}
			}
			if(k==0)
			sum+=p[i].val;
		}
		printf("%I64d\n",sum);
	}
	return 0;
}